# Lagrange identity for determinants

Let A $$\in M_{(n-1)},n(\mathbb{R})$$ and for each $$1\leq j \leq n$$,let $$A_j$$ the matrix obtained from A by removing the j-th column. Show that:

$$det (AA^t)= \sum\limits_{j=1}^n det(A_j)^2$$

My first thought was using Laplace expansion,but this exercise is in the section of symmetric and exterior powers of my book. So I think maybe I should use the fact that the determinant is an alternating form . Any hints on how to proceed?

Consider $$\det\pmatrix{x^t\\A}$$, where $$x\in \mathbb{R}^n$$.

1) Notice that $$\pmatrix{r_j^t\\A}$$ has two equal rows for every row $$r_j^t$$ of $$A$$. Thus, $$\det\pmatrix{r_j^t\\A}=0$$ .

2) Expanding $$\det\pmatrix{x^t\\A}$$ using Laplace on the first row, we get $$\det\pmatrix{x^t\\A}=x^tv,$$ where $$v^t=((-1)^{1+1}\det(A_1),\ldots,(-1)^{1+n}\det(A_n)).$$

By item 1) and 2), $$r_j^tv=0$$ for every row $$r_j$$ of $$A$$. Therefore, $$\pmatrix{v^t\\A}\pmatrix{v^t\\A}^t=\pmatrix{v^tv & 0\\0 & AA^t}$$.

Therefore, $$v^tv\ \det(AA^t)=\det\left(\pmatrix{v^t\\A}\pmatrix{v^t\\A}^t\right)=\det\left(\pmatrix{v^t\\A}\right)^2=(v^tv)^2.$$

If $$v\neq 0$$ then $$\det(AA^t)=v^tv=\sum_{i=1}^n\det(A_i)^2$$.

If $$v=0$$ then let $$y^t\neq 0$$ be any row vector in the orthogonal complement of $$span\{r_1,\ldots,r_{n-1}\}$$.

So $$\pmatrix{y^t\\A}\pmatrix{y^t\\A}^t=\pmatrix{y^ty & 0\\0 & AA^t}$$ and

$$y^ty\ \det(AA^t)=\det\left(\pmatrix{y^t\\A}\pmatrix{y^t\\A}^t\right)=\det\left(\pmatrix{y^t\\A}\right)^2=(y^tv)^2=0\ (\text{since } v=0).$$

Since $$y\neq 0$$ then $$y^ty\neq 0$$ and $$\det(AA^t)=0=v^tv=\sum_{i=1}^n\det(A_i)^2$$.

In both cases, $$v=0$$ or $$v\neq 0$$, we get $$\det(AA^t)=\sum_{i=1}^n\det(A_i)^2$$.